(1) Field of the Invention
The present invention relates to a precision current mirror circuit, and more particularly, to a mirror circuit which eliminates current loss on a mirror side, and ultimately causes both the current source and output current to be the same.
(2) Description of the Prior Art
A current mirror circuit has two transistors with their base terminals connected to each other. Both transistors are identical; that is, they are fabricated to have matched characteristics so that the same amount of current flows in the two transistors. The current mirror thereby controls a current in one transistor so as to also control a current in the other transistor. With such characteristics, the current mirror is widely used for analog circuits.
FIG. 1 shows a conventional current mirror circuit.
As shown in FIG. 1, the conventional current mirror circuit includes a first transistor Q1 for receiving a supply voltage V.sub.cc through an emitter, and having its base terminal, connected to its collector; a second transistor Q2 for receiving the supply voltage V.sub.cc through the emitter, and having the base terminal of the first transistor Q1 connected to its own base terminal; and a current source I.sub.in having an input for receiving the collector of the first transistor Q1, and having an output terminal grounded.
After the conventional current mirror circuit receives the supply voltage V.sub.cc, since its collector and base are connected to each other, the first transistor Q1 achieves a diode connection to the current source I.sub.in, and the current starts to flow.
Since each of the transistors Q1 and Q2 has a base current I.sub.B, the current I.sub.Q1 through the transistor Q1 decreases by as much as the base currents of the transistors, i.e., 2I.sub.B, as shown in the following expression (1). EQU I.sub.Q1 =I.sub.in -2I.sub.B ( 1)
The current I.sub.Q2 (i.e., I.sub.out1) through the second transistor Q2, being related to the first transistor Q1 in terms of the current mirror, also decreases as much as the base currents of the transistors, i.e., 2I.sub.B, as shown in the following expression (2). EQU I.sub.out1 =I.sub.in -2I.sub.B ( 2)
That is, the conventional current mirror has a problem in that the output current I.sub.out1 of the mirror side decreases as much as the base currents 2I.sub.B of the transistors Q1 and Q2 as compared with the input current I.sub.in. A circuit for reducing the above-identified problem is shown in FIG. 2.
FIG. 2 is a conventional current mirror circuit for reducing the problem of the circuit of FIG. 1, and includes a first transistor Q1 for receiving a supply voltage V.sub.cc through its emitter; a second transistor Q2 for receiving the supply voltage V.sub.cc through its emitter, and having the base terminal of the first transistor Q1 to its own base terminal; a current source I.sub.in having an input for receiving the collector of the first transistor Q1, and having an output terminal grounded; and a third transistor Q3 having the base terminal of the first transistor Q1 connected to its own emitter terminal, and having the collector terminal of the first transistor Q1 connected to its own base terminal, and having its collector terminal grounded.
Since the first transistor Q1 is connected to the current source I.sub.in when the supply voltage V.sub.cc is applied to the circuit, the current starts to flow.
Because each of the transistors Q1 and Q2 has a base current I.sub.B, the current 2I.sub.B corresponding to the base currents of the transistors Q1 and Q2 flows through the emitter of the third transistor Q3. The base current I.sub.Q3B of the third transistor Q3 is shown in the following expression (3). EQU I.sub.Q3B =2I.sub.B /H.sub.FE ( 3)
Consequently, the current I.sub.Q1 through the first transistor Q1, as shown in the following expression (4), decreases by as much as the base current I.sub.Q3B of the third transistor Q3. EQU I.sub.Q1 =I.sub.in -2I.sub.B /H.sub.FE ( 4)
Accordingly, current I.sub.Q2 through the collector of the second transistor Q2, being related to the first transistor Q1 in terms of the current mirror, decreases by as much as the base current I.sub.Q3B of the third transistor Q3. The output current I.sub.out2 is shown in the following expression (5). EQU I.sub.out2 =I.sub.in -2I.sub.B /H.sub.FE ( 5)
It should be noted that even if the conventional current mirror circuit in FIG. 2 has an improved output current compared with the circuit of FIG. 1, it still has a problem in that the output current I.sub.out2 of the mirror side decreases by as much as the base current I.sub.Q3B of the third transistor Q3 as compared with the input current I.sub.in.